Division euclidienne

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(X−3)2n+ (X−2)n−2 (X−2)2
A∈K[X]>0P∈Kn[X]
P0, P1, . . . , Pn
(deg Pi<deg A
P=P0+P1A+··· +PnAn.
B∈K[X]n > 0Φ : K[X]−→ Kn−1[X]
P7−→ R
Ψ : K[X]−→ K[X]
P7−→ QP=QB +R
Φ Ψ
Φ(P1P2)
P7−→ AP mod B
E=K3[X], A =X4−1, B =X4−X, ϕ:E−→ E
P7−→ AP B.
ϕ ϕ
P∈K[X]a, b ∈Kα=P(a)β=P(b)
P(X−a)(X−b)
(cos θ+Xsin θ)nX2+ 1
P∈Q3[X]X+ 1 X+
2, X + 3, X + 4
P Q
P=X4+X3−3X2−4X−1
Q=X3+X2−X−1
P=X4−10X2+ 1
Q=X4−4X3+ 6X2−4X+ 1
P=X5−iX4+X3−X2+iX −1
Q=X4−iX3+ 3X2−2iX + 2
P Q U, V ∈K[X]
UP +V Q = 1
P=X4+X3−2X+ 1
Q=X2+X+ 1
P=X3+X2+ 1
Q=X3+X+ 1

(X+ 1)n−Xn−1X2+X+ 1
(X+ 1)n−Xn−1X2+X+ 1
n∈N(1 + X4)n−Xn1 + X+X2R[X]
(X−2)2n+ (X−1)n−1 (X−1)(X−2)
Pn= (X−2)2n+ (X−1)n−1
PnX−1X−2Q1Q2
Pn(X−1)(X−2) Q2−Q1
(X−2)2n−2−(X−2)2n−3+··· − (X−2) + 1+(X−1)n−2+ (X−1)n−3+··· + (X−1) + 1.
X50 X2−3X+ 2
X+√317 X2+ 1
X8−32X2+ 48 X−√23
λ, µ ∈CX2+X+ 1 X5+λX3+µX2+ 1
P∈K[X]P X2+ 1 X2−1 2X−2
−4X P X4−1
(Xn−1, Xm−1)
m, n ∈N∗(Xn−1, Xm−1)
P, Q ∈K[X]D= (P, Q)
U, V ∈K[X]
UP +V Q =D
deg U < deg Q−deg D
deg V < deg P−deg D.
U V
(U, V )7−→ UA +V B
A, B ∈K[X]p= deg A q = deg BΦ : Kq−1[X]×Kp−1[X]−→ Kp+q−1[X]
(U, V )7−→ UA +V B
A∧B= 1 ⇐⇒ Φ
(P(X), P (−X)) (P(X), P (−X))
P∈K[X] (P(X), P (−X)) (P(X), P (−X))
A◦P|B◦P⇒A|B
A, B, P ∈K[X]P A ◦P B ◦P A B

−2nX + 4n−1
ϕ={P∈E X −1|P}
ϕ= (X3+X2+X)
α(b−X) + β(X−a)
b−a
cos nθ +Xsin nθ
P=λ((X+ 2)(X+ 3)(X+ 4) −6)
X+ 1
1
X2−iX + 1
7U=X+ 3,7V=−X3−3X2+X+ 4
3U= 2X2−X+ 1,3V=−2X2−X+ 2
j X ⇒R=
(−1)n−2n≡0 [3]
((−1)n+1 −1)(X+ 1) n≡1 [3]
((−1)n+ 1)X n ≡2 [3].
n≡0 [6]
(250 −1)X+ 2 −250
216X−√3
192X−√22
λ=µ=−1
−3X3+X2−X−1
n=qm +r⇒Xn−1≡Xr−1 [Xm−1] n
m⇒=Xn∧m−1